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...default in R is formed using the so-called "treatment" contrasts. To wander from R into statistics (sorry Bert) the problem arises because the "usual" parametrisation of the model is the "over-parametrised" form: Y_ij = mu + beta_i + E_ij (i = 1, ..., I, j = 1, ..., J_i) where Y_ij is the j-th observation corresponding to the i-th "treatment" or group. (Things get a bit more complicated in "multi-way" models; let's not go there.) The parameter "mu" is the "grand mean" and the "beta_i" are the "treatm...

This is really a statistical issue. What do you think the Intercept term represents? See ?contrasts. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Thu, Feb 15, 2018 at 5:27 PM, Marc Girondot via R-help < r-help at

Dear friends, two years ago (as I found on the web) Paul sent the following message but I was not able to find if he got an answer. Today I have the same question and it would be great if I could find out that this test has been implemented (somehow) in R. Please do not confuse it with the Brown-Forsythe test of equality of variances. Thank you: I've been searching around for a function for

Dear R-er, I try to get the standard error of fitted parameters for factors with a glm, even the reference one: a <- runif(100) b <- sample(x=c("0", "1", "2"), size=100, replace = TRUE) df <- data.frame(A=a, B=b, stringsAsFactors = FALSE) g <- glm(a ~ b, data=df) summary(g)$coefficients # I don't get SE for the reference factor, here 0:

I read somewhere that t.test is equivalent to a hypothesis testing for one way ANOVA. But I'm wondering how they are equivalent. In the following code, the p-value by t.test() is not the same from the value in the last command. Could somebody let me know where I am wrong? > set.seed(0) > N1=10 > N2=10 > x=rnorm(N1) > y=rnorm(N2) > t.test(x,y) Welch Two Sample t-test data: